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標題: math [打印本頁]

作者: questerli 時間: 7/11/2005 11:26 AM
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作者: questerli 時間: 7/11/2005 10:08 PM
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作者: 宇文水靈 時間: 8/11/2005 09:40 PM
b(i). Since the coefficient of x^2 is 2,
>0
∴The curve opens down.
b(ii)f(x)=2x^2+3x+1
   =2[x^2+3/2x]+1
   =2[(x+3/4)^2-(3/4)^2]+1
   =2(x+3/4)^2-1/8
Line of symmetery= -3/4
(Alternative)
Line of symmetery= -b/2a
                     = -3/4
b(iii)At X-intercept, y=0
   y=2x^2+3x+1
   =(2x+1)(x+1)
X-intercept = -1/2,-1
b(iv)f(x)=2x^2+3x+1
         =2(x+3/4)^2-1/8 By b(ii)
Vertex (-3/4, -1/8)
Since the curve opens up, the vertex is minimum point.
c./懶了，用Graphmatica畫俾你算了，但係咁樣會提高準確度。

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[ Last edited by 宇文水靈 on 2005-11-8 at 09:57 PM ]

作者: questerli 時間: 9/11/2005 04:03 PM
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